Question: The altitude of an equilateral triangle is $\sqrt6$ units.  What is the area of the triangle, in square units? Express your answer in simplest radical form.
Solution: Drawing an altitude of an equilateral triangle splits it into two 30-60-90 right triangles: [asy]
unitsize(0.6inch);
pair A, B, C, F;
A = (0,1);
B = rotate(120)*A;
C = rotate(120)*B;
F = foot(A,B,C);
draw(A--B--C--A,linewidth(1));
draw(A--F);
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$M$",F,S);
[/asy]


The altitude is the longer leg of each 30-60-90 triangle, so $AM = \sqrt{3}\cdot BM$ in the diagram above. Since $AM = \sqrt{6}$, we have \[BM = \frac{AM}{\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac63} = \sqrt{2}.\] Therefore, we have $BC = 2BM = 2\sqrt{2}$, so the area of the triangle is $(BC)(AM)/2 = (2\sqrt{2})(\sqrt{6})/2
=\sqrt{12} = \boxed{2\sqrt{3}}$ square units.